ГДЗ по алгебре 7 класс Макарычев ФГОС Задание 819

\[\boxed{\text{819.}\text{\ }еуроки - ответы\ на\ пятёрку}\]

\[\textbf{а)}\ (2x + 3)^{2} = (2x)^{2} + 2 \cdot 2x \cdot\]

\[\cdot 3 + 3² = 4x² + 12x + 9\]

\[\textbf{б)}\ (7y - 6)^{2} = (7y)^{2} - 2 \cdot 7 \cdot\]

\[\cdot 6y + 6^{2} = 49y^{2} - 84y + 36\]

\[\textbf{в)}\ (10 + 8k)^{2} = 10² + 8 \cdot 10 \cdot\]

\[\cdot 2k + (8k)^{2} = 100 + 160k +\]

\[+ 64k²\]

\[\textbf{г)}\ (5y - 4x)^{2} = (5y)^{2} - 2 \cdot 5 \cdot\]

\[\cdot 4yx + (4x)^{2} = 25y^{2} -\]

\[- 40yx + 16x²\]

\[\textbf{д)}\ \left( 5a + \frac{1}{5}b \right)^{2} = (5a)^{2} + 2 \cdot 5 \cdot\]

\[\cdot \frac{1}{5}ab + \left( \frac{1}{5}b \right)^{2} = 25a^{2} +\]

\[+ 2ab + \frac{1}{25}b²\]

\[\textbf{е)}\ \left( \frac{1}{4}m - 2n \right)^{2} = \left( \frac{1}{4}m \right)^{2} - 2 \cdot\]

\[\cdot 2 \cdot \frac{1}{4}mn + (2n)^{2} =\]

\[= \frac{1}{16}m^{2} - mn + 4n²\]

\[\textbf{ж)}\ (0,3x - 0,5a)^{2} = (0,3x)^{2} -\]

\[- 2 \cdot 0,3 \cdot 0,5xa + (0,5a)^{2} =\]

\[= 0,09x^{2} - 0,3ax + 0,25a²\]

\[\textbf{з)}\ (10c + 0,1y)^{2} = (10c)^{2} + 2 \cdot\]

\[\cdot 10 \cdot 0,1cy + (0,1y)^{2} =\]

\[= 100c² + 2cy + 0,01y²\ \]

\[\textbf{и)}\ (0,1b - 10a)^{2} = (0,1b)^{2} - 2 \cdot\]

\[\cdot 0,1b \cdot 10a + (10a)^{2} =\]

\[= 0,01b^{2} - 2ab + 100a^{2}\]